Suppose $X$ is a reflexive space (possibly non-separable) which is not super-reflexive. Then (by definition) there exists a non-reflexive Banach space $Y$ which is non-reflexive but is finitely representable in $X$, meaning that for each $\lambda >1$, every finite dimensional subspace of $Y$ is $\lambda$-isomorphic to a subspace of $X$. Can we always find such $Y$ (i.e. non-reflexive) which is separable? In this spirit, what are examples of reflexive but not super-reflexive spaces in which neither $\ell_1$ nor $c_0$ is finitely representable?

What does the existence of (twice!) non-reflexive $Y$ have to do with any properties of $X$? There are certainly non-reflexive separable spaces … but I suspect that gremlins ate half the text of your question.
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Harald Hanche-OlsenJan 4 '13 at 20:37

1 Answer
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The first question is easy: Every non reflexive space has a separable non reflexive subspace (e.g. by the Eberlein-Smulian theorem or by R. C. James' characterization of non reflexivity).

The second question was a longstanding open problem that was solved by James in the 1970s. Pisier and Xu gave another proof--you can find their paper by using MathSciNet. Their approach is more conceptual and uses interpolation theory but is not easy.

Dear Prof. Johnson. Thank you. I've been trying to find the papers with no success yet, but I'll try again. Let me ask then whether the answer to the second question is positive or negative. :)
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Bojan KwitekJan 4 '13 at 22:34

OK, thank you. I haven't spotted this paper. By the way, can we deduce from the fact $\ell_1$ is finitely representable in $X$ that $c_0$ is finitely representable in $X^*$?
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Bojan KwitekJan 6 '13 at 22:56

No, Bojan--$X=c_0$ is a counterexample. You do get that $\ell_1$ is finitely representable in $X^*$. Having $c_0$ finitely representable in $X^*$ is equivalent to $X$ containing uniformly complemented uniform copies of $\ell_1^n$.
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Bill JohnsonJan 28 '13 at 17:36